PX153 - C2 - construction of ODEs

eg: you have lump of radioactive material. how does the number of radioactive nuclei change over time?

• variables:
  • independent - time($t$)
  • dependent - number of radio active nuclei ($N(t)$)
• empirical observation:
  • rate of decay is proportional to the number of nuclei: $\frac{dN}{dt}=-\alpha N$
• try a solution: $N(t)=A{e}^{\lambda t}$
  $\frac{dN}{dt}=\lambda A{e}^{\lambda t}=-\alpha N$
  $\lambda =-\alpha$
• and the general solution: $N(t)=A{e}^{-\alpha t}$
• aside: why $N(t)=A{e}^{-\alpha t}$?
  • $e^x$ is defined by, if $y=e^x$, $y=y^{\prime }$, $y(0)=1$
  • if $\frac{dN}{dt}\propto N$, solution will be an exponential
• to find a particular solution, we need a boundary condition
  • if $N(0)=N_0$ then, the particular solution is $N(t)=N_0{e}^{-\alpha t}$

eg: a large tank of water with cross-section $A$, and a small opening on the side, at the bottom, with cross-section $a$. find the ODE that described the height of the water

• 
• variables:
  • independent - $t$
  • dependent - $h(t)$
• the water exits the opening at speed $v(h)$
  • in a time $\mathrm{\Delta }t$, the volume of water leaving is $\mathrm{\Delta }V=av(h)\mathrm{\Delta }t$
    • corresponds to a change in height:
    $$\mathrm{\Delta }h=h(t+\mathrm{\Delta }t)-h(t)=\frac{\mathrm{\Delta }v}{A}=\frac{-av(h)\mathrm{\Delta }t}{A}$$
    • to find $v(h)$, we use bernaulli's equation of energy conservation in fluid flow:
    $$p_0+\rho gh+\frac{1}{2}\rho v^2=constant$$- basically $E=KE+PE$$$⟹\rho gh=\frac{1}{2}\rho v^2⟹v(h)=\sqrt{2gh}$$
    • so we have
    $$\frac{h(t+\mathrm{\Delta }t)-h(t)}{\mathrm{\Delta }t}=-\frac{a}{A}\sqrt{2gh}$$
    • taking the limit $\mathrm{\Delta }t\to 0$ gives is the derivative
    $$\frac{dh}{dt}=-\frac{a}{A}\sqrt{2gh}$$